Question: Simplify and expand the following expression: $ \dfrac{q + 3}{3q - 8}-\dfrac{q}{q + 7} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3q - 8)(q + 7)$ Multiply the first term by $\dfrac{q + 7}{q + 7}$ $ \begin{align*} \dfrac{q + 3}{3q - 8} \times \dfrac{q + 7}{q + 7} & = \dfrac{(q + 3)(q + 7)}{(3q - 8)(q + 7)} \\ & = \dfrac{q^2 + 10q + 21}{(3q - 8)(q + 7)}\end{align*} $ Multiply the second term by $\dfrac{3q - 8}{3q - 8}$ $ \begin{align*} \dfrac{q}{q + 7} \times \dfrac{3q - 8}{3q - 8} & = \dfrac{(q)(3q - 8)}{(q + 7)(3q - 8)} \\ & = \dfrac{3q^2 - 8q}{(q + 7)(3q - 8)}\end{align*} $ Now we have: $ = \dfrac{q^2 + 10q + 21}{(3q - 8)(q + 7)} - \dfrac{3q^2 - 8q}{(q + 7)(3q - 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{q^2 + 10q + 21 - (3q^2 - 8q)}{(3q - 8)(q + 7)} $ $ = \dfrac{q^2 + 10q + 21 - 3q^2 + 8q}{(3q - 8)(q + 7)} $ $ = \dfrac{-2q^2 + 18q + 21}{(3q - 8)(q + 7)}$ Expand the denominator: $ = \dfrac{-2q^2 + 18q + 21}{3q^2 + 13q - 56}$